Solve p^4-13p^2+36=0

$$p ^ { 4 } - 13 p ^ { 2 } + 36 = 0$$

Solve for p

$$p=-3$$

$$p=3$$

$$p=2$$

$$p=-2$$

Steps Using the Quadratic Formula

Substitute t for p^{2}.

$$t^{2}-13t+36=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -13 for b, and 36 for c in the quadratic formula.

$$t=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\times 1\times 36}}{2}$$

Do the calculations.

$$t=\frac{13±5}{2}$$

Solve the equation t=\frac{13±5}{2} when ± is plus and when ± is minus.

$$t=9 t=4$$

Since p=t^{2}, the solutions are obtained by evaluating p=±\sqrt{t} for each t.

$$p=3 p=-3 p=2 p=-2$$